【01背包】HDU 2602 Bone Collector （模板题）

Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output

One integer per line representing the maximum of the total value (this number will be less than 2³¹).

Sample Input

1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output

14

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

 

思路：01背包裸题。01背包：有限的背包容量取最大的价值，每样东西选择取或不取

 

 #include <iostream>
 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 using namespace std;

 long long va[];
 int vo[];
 long long dp[];

 int main()
 {
     int T,n,v,i,j;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&v);
         for(i=;i<=n;i++)
             scanf("%lld",&va[i]);
         for(i=;i<=n;i++)
             scanf("%d",&vo[i]);
         memset(dp,,sizeof(dp));
         for(i=;i<=n;i++)
         {
             for(j=v;j>=vo[i];j--)
             {
                 dp[j]=max(dp[j],dp[j-vo[i]]+va[i]);
             }
         }
         printf("%lld\n",dp[v]);
     }
     return ;
 }